Aufgaben:Exercise 1.1Z: Low-Pass Filter of 1st and 2nd Order

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Damping– & phase function

The simplest form of a low-pass filter – for example, realisable as an RC low-pass filter according to  exercise 1.1  – has the following frequency response:

$$H_{\rm 1}(f) = \frac{1}{1+{\rm j}\cdot f/f_0}.$$

This is then referred to as a low-pass filter of first order. The diagram shows the following for this filter

• above the damping curve  $a_1(f)$,
• below the phase response  $b_1(f)$.

Correspondingly, for a low-pass filter of  $n$–th order the following defining equation applies:

$$H_n(f) = H_{\rm 1}(f)^n.$$

In this task,

• based on the functions  $a_1(f)$  and  $b_1(f)$  of the low-pass filter of first order
• the damping curve and phase response of a low-pass filter of higher order is to be analysed.

In general, the following holds:

$$H(f) = {\rm e}^{-a(f) - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}.$$

Please note:

• The task belongs to the chapter  System Description in Frequency Domain.

• There is the following relationship between the Np- and dB-values of an amplitude value  $|H| = 1/x$ :

$$a_{\rm Np} = \ln (x) = \ln (10) \cdot \lg (x) = \frac{\ln (10)}{20} \cdot a_{\rm dB} \approx 0.11513 \cdot a_{\rm dB}.$$

• Further consider that for two complex quantities  $z_1$  and  $z_2$  the following equations hold:

$$|z_1 \cdot z_2| = |z_1| \cdot |z_2|, \hspace{0.5 cm}{\rm arc}\hspace{0.05 cm}(z_1 \cdot z_2) = {\rm arc}\hspace{0.05 cm}(z_1) + {\rm arc}\hspace{0.05 cm}(z_2).$$

Questions

1

Compute the damping curve  $a_1(f)$  of a low-pass filter of first order in  $\rm dB$.
What  $\rm dB$–values are obtained for  $f = f_0$  and  $f = 2f_0$?

$a_1(f = f_0)\ = \ $  $\text{dB}$

$a_1(f = 2f_0)\ = \ $  $\text{dB}$

2

Compute the phase response  $b_1(f)$.
What values in radian (rad) are obtained for  $f = f_0$  and  $f = 2f_0$?

$b_1(f = f_0)\ = \ $  $\text{rad}$

$b_1(f = 2f_0)\ = \ $  $\text{rad}$

3

What is the damping curve  $a_n(f)$  of a low-pass filter of  $n$–th order?
What  $\rm dB$–values are obtained for  $n = 2$  for  $f = f_0$  or  $f = \: –2f_0$?

$a_2(f = f_0)\ = \ $  $\text{dB}$

$a_2(f = -2f_0)\ = \ $  $\text{dB}$

4

Compute the phase function  $b_2(f)$  of a low-pass filter of second order.
What values (in radian) are obtained for  $f = f_0$  and  $f = \: –2f_0$?

$b_2(f = f_0)\ = \ $  $\text{rad}$

$b_2(f = -2f_0)\ = \ $  $\text{rad}$

Sample solution

Solution

(1)  The amplitude response of the low-pass filter of first order is:

$$|H_{\rm 1}(f)| = \frac{1}{\sqrt{1+ (f/f_0)^2}}.$$

• This gives the damping curve in Neper (Np):

$$a_1(f) = \ln \frac{1}{|H_1(f)|} = {1}/{2} \cdot \ln \left[1 + ({f}/{f_0})^2 \right] \Rightarrow a_1(f = f_0) = 0.3466 \hspace{0.1 cm}{\rm Np},\hspace{0.5 cm}a_1(f = 2 f_0) = 0.8047 \hspace{0.1 cm}{\rm Np}.$$

The corresponding dB-values are obtained by multiplying by  $1/0.11513 = 8.68589$  and result in

• $ \underline{3.01 \: {\rm dB} ≈ 3 \: {\rm dB}}$  für  $ f = f_0$,
• $ \underline{6.99 \: {\rm dB}}$  für  $ f = 2f_0$.

Thus, for the low-pass filter of first order the 3dB–cutoff frequency is   $f_{\rm G} = f_0$.

(2)  The frequency response  $H_1(f)$  can also be represented separately by real and imaginary parts:

$$H_{\rm 1}(f) = \frac{1}{ {1+ (f/f_0)^2} } - {\rm j} \cdot \frac{f/f_0}{ {1+ (f/f_0)^2} }.$$

• Hence, the following holds for the phase response:

$$b_1(f) = - \arctan \hspace{0.1cm} ( {\rm Im} /{\rm Re} ) = \arctan \hspace{0.1cm} ({f}/{f_0}).$$

• The value  $\arctan(1) = π/4 \rm \underline{\: = 0.786 \: rad}$ is obtained for  $f = f_0$  and   $\arctan(2) \rm \underline{\: = 1.108 \: rad}$ for  $f = 2f_0$ .

(3)  For the amplitude response of a low-pass filter of  $n$–th order the following is valid:

$$|H_n(f)| = |H_{\rm 1}(f)|^n.$$

Regarding the (logarithmic) damping function the  $n$–fold multiplication becomes a   $n$–fold sum:

$$a_n(f) = n \cdot a_1(f)= {n}/{2} \cdot \ln \left[ 1 + ({f}/{f_0})^2 \right].$$

For the low-pass filter of second-order this results in the special case:

$$a_2(f) = \ln \left[ 1 + ({f}/{f_0})^2 \right]= 2 \cdot a_1(f).$$

The dB-values are now:

• $ \underline{6.02 \: {\rm dB} ≈ 6 \: {\rm dB}}$  für  $f = ±f_0$,
• $\rm \underline{13.98 \: {\rm dB} ≈ 14 \: {\rm dB}}$  für  $f = ±2f_0$.

It is thus obvious that for  $n > 1$  the parameter  $f_0$  no longer indicates the 3dB-cutoff frequency $f_{\rm G}$.
For  $n = 2$   ⇒   "low-pass filter of second order" the following relationship applies instead:  

$${f_{\rm G} } = {f_0}/\sqrt{2}.$$

(4)  Also, with respect to the phase function the following holds:

$$b_n(f) = n \cdot b_1(f), \hspace{0.3 cm} b_2(f) = 2 \cdot b_1(f).$$

Thus, for the low-pass filter of second order all phase values between  $±π$  are possible. In particular the following holds:

• $b_2(f = f_0) = π/2 \rm \underline{\: = 1.571 \: rad}$,
• $b_2(f = 2f_0) = \rm 2.216 \: rad$.

Since phase is an odd function the following applies to it:   $b_2(f = \: –2f_0) = \rm \underline{–2.216 \: rad}$.